Let f(x) = log(x2 + 2) – log3
Clearly f is continous on [–1, 1] and
f is derivable on (–1, 1)
Also, f(–1) = log (1 + 2) – log 3 = log 3 – log 3 = 0
f(1) = log (1 + 2) – log 3 = log 3 – log 3 = 0
∴ f(–1) = f(1)
f satisfies all the conditions of Rolle's theorem.
∴ There exists C ∈ (–1, 1) such that f1 (c) = 0
But f(x) = log (x2 + 2) – log 3
2c = 0
⇒ c = 0 ∈ (–1, 1)
Hence Rolle's theorem is verified.